A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.

1.

2 cm

2.

3 cm

3.

1 cm

4.

3.5 cm

5.

None of these

Solution

Answer:

Volume of sphere = \(\frac{4}{3}\pi {r^3}\)

Where, r is radius of sphere.

Volume of the given sphere = (4/3) × π × 63

This volume is converted to the hollow cylinder.

Let, the thickness of cylinder be x.

External radius = 5 cm

Internal radius = (5 – x) cm

Area of circular base = πr2, where r is radius of base.

Area of hollow base = (π × 52) – π (5 – x)2

= π [52 – (5 – x)2]

= π (5 + 5 – x) (5 – 5 + x)

= πx(10 – x)

Volume of cylinder = Area of base × height

∴(4/3) × π × 63 = 32 × π (10 – x)x

⇒ 9 = 10x – x2

⇒ x2 – 10x + 9 = 0

⇒ x2 – 9x – x + 9 = 0

⇒ x(x – 9) – 1(x – 9) = 0

⇒ (x – 1) (x – 9) = 0

⇒ x = 1 or x = 9

x = 9 not possible, since the thickness cannot be greater than 5, which is the external radius.